Answer:
10 questions with an incorrect answer for both parts
15 questions with a correct answer in only one part
Explanation:
Let's call x the number of questions with an incorrect answer for both parts and y the number of questions with only one part of the question correct.
So, the equation that gives us the number of points after 25 rounds is:
100 - x - 0.5y = 82.5
Where: x + y = 25
So, solving for y, we get:
y = 25 - x
Replacing this on the initial equation and solving for x, we get:
[tex]\begin{gathered} 100-x-0.5y=82.5 \\ 100-x-0.5(25-x)=82.5 \\ 100-x-12.5+0.5x=82.5 \\ 87.5-0.5x=82.5 \\ 87.5-82.5=0.5x \\ 5=0.5x \\ \frac{5}{0.5}=x \\ 10=x \end{gathered}[/tex]Then, the value of y is:
[tex]\begin{gathered} y=25-x \\ y=25-10 \\ y=15 \end{gathered}[/tex]Therefore, the player gets 10 questions with an incorrect answer for both parts and 15 questions with a correct answer in only one part.
What are all the factors of 54?
To find the factors of a number, we can look for factors to divide it by subsequently.
It is easier to start with lower factors.
Let's start by "2".
Since "54" is even, it is divisable by "2":
[tex]\frac{54}{2}=27[/tex]So "2" is one of the factors.
Now, we have got 27. It is not even anymore, but it is divisable by "3":
[tex]\frac{27}{3}=9[/tex]So "3" is another factor.
Now we have got "9" and it is also divisable by "3":
[tex]\frac{9}{3}=3[/tex]So there is another "3" factor.
And since we have got now another "3", we know it is divisable by "3":
[tex]\frac{3}{3}=1[/tex]Now we have got to "1", so we found all the prime factors:
[tex]54=2\cdot3\cdot3\cdot3[/tex]Now,, we need to combine them to find all possible combinations.
We will start from low to high.
"1" is always a factor.
There is "2" there, so it is also a factor.
Then we have "3" as another factor.
There is no need to combine "1" with another factor, so we will start b combining 2 and 3:
[tex]2\cdot3=6[/tex]So, "6" is another factor.
We can combine 3 with 3:
[tex]3\cdot3=9[/tex]"9" is another factor.
Now we start combining three of them:
[tex]\begin{gathered} 2\cdot3\cdot3=6\cdot3=18 \\ 3\cdot3\cdot3=9\cdot3=27 \end{gathered}[/tex]So, "18" and "27" are factors.
And now we combine 4 of them, but this is get us back to "54" which is the last factor.
So, the factors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54.
Answer:
Step-by-step explanation:
1 × 54 = 54, (1, 54) is a pair factor of 54.
2 × 27 = 54, (2, 27) is a pair factor of 54
3 × 18 = 54, (3, 18) is a pair factor of 54
6 × 9 = 54, (6, 9) is a pair factor of 54
Therefore, the positive pair factors are (1, 54), (2, 27),(3, 18) and (6, 9).
geometry extra credit- 25 questions
Given that the triangles are similar, then their corresponding sides are in proportion, that is,
[tex]\frac{AB}{JK}=\frac{BX}{KY}[/tex]Substituting with data and solving for BX:
[tex]\begin{gathered} \frac{32}{10}=\frac{BX}{6} \\ 3.2=\frac{BX}{6} \\ \text{3}.2\cdot6=BX \\ 19.2=BX \end{gathered}[/tex]I need answer for this word problems you have to shown that you can make several lattes then you add milk and begin to stirring. you use a total of 30 ounces of liquid. write an equation that represents the situation and explain what the variable represents.
hello
the question here is a word problem and we can either use alphabhets to represent the variables.
let lattes be represented by x and milk be represented by y
[tex]x+y=30[/tex]since the total ounce of liquid is equals to 30, we equate the whole sentence to 30.
To the nearest whole foot, how many feet would it be to walk diagonally across this field? A. 42B. 50C. 65D. None of the above
Can you please help with 44For the following exercise, sketch a graph of the hyperbola, labeling vertices and foci
We have the following equation of a hyperbola:
[tex]4x^2+16x-4y^2+16y+16=0[/tex]Let's divide all the equations by 4, just to simplify it
[tex]x^2+4x-y^2+4y+4=0[/tex]Just to make it easier, let's put the term if "x" isolated
[tex]x^2+4x=y^2-4y-4[/tex]Now we can complete squares on both sides, just remember that
[tex]\begin{gathered} (a+b)^2=a^2+2ab+b^2 \\ \\ (a-b)^2=a^2-2ab+b^2 \end{gathered}[/tex]Now let's complete it!
[tex]\begin{gathered} x^2+4x=y^2-4y-4\text{ complete adding 4 on both sides} \\ \\ x^2+4x+4=y^2-4y-4+4 \\ \\ (x+2)^2=y^2-4y \\ \end{gathered}[/tex]We already completed one side, now let's complete the side with y^2, see that we will add 4 again, then
[tex]\begin{gathered} (x+2)^2=y^2-4y \\ \\ (x+2)^2+4=y^2-4y+4 \\ \\ (x+2)^2+4=(y-2)^2 \end{gathered}[/tex]And now we can write it using the standard equation!
[tex]\begin{gathered} (y-2)^2-(x+2)^2=4 \\ \\ (y-2)^2-(x+2)^2=4 \\ \\ \frac{(y-2)^2}{4}-\frac{(x+2)^2}{4}=1 \end{gathered}[/tex]And now we can graph it like all other hyperbolas, the vertices will be:
[tex](-2,4)\text{ and }(-2,0)[/tex]And the foci
[tex]\begin{gathered} c^2=a^2+b^2 \\ \\ c^2=2^2+2^2 \\ \\ c^2=2\cdot2^2 \\ \\ c^{}=2\, \sqrt[]{2} \end{gathered}[/tex]Then the foci are
[tex](-2,2+2\, \sqrt[]{2})\text{ and }(-2,2-2\, \sqrt[]{2})[/tex]Now we can plot the hyperbola!
Factoring
64v^4-225w^10
I come up with
(8v^2+15w^2)(8v^2-15w^5) can I simplify it?
Your answer can't be simplified but it's also slightly incorrect. No worries!
You can simplify (64v^4 - 225w^10) by applying the difference of squares --> a^2 - b^2 = (a + b)(a - b)
Here, (64v^4) is a^2, and (225w^10) is b^2.
[tex]\sqrt{64v^4 }[/tex] = a = 8v^2
[tex]\sqrt{225w^{10} }[/tex] = b = 15w^5
Substitute the values:
(8v^2 + 15w^5)(8v^2 - 15w^5)
Therefore, the factorization of [tex]64x^{4} -225w^{10}[/tex] is:
[tex](8v^{2} + 15w^5)(8v^2-15w^5)[/tex]
The factorization is already the simplest. It can't be simplified further.
4j+2=h solve for j please help
To solve for j'
4j + 2 = h
subtract 2 from both-side of the equation
4j + 2-2 = h -2
4j = h-2
divide both-side of the equation by 4
4j/4 = h-2/4
[tex]j=\frac{h-2}{4}[/tex]Find the perimeter of LMNPQ if the perimeter of ABCDE = 25.2 cm and ABCDE = LMNPQ
If two polygons are similar with the lengths of corresponding sides, we need to find the ratio between both figures:
Then:
[tex]\frac{SIDE\text{ PQ}}{\text{SIDE DE}}=\frac{25}{15}=\frac{5}{3}=\text{ 1,666}[/tex]So, the ratio is 5:3
The perimeter of LMNPQ is the same to say :
P of ABCDE = 25.2cm
Use the ratio
P of LMNPQ = 25.2 (5/3) cm
Then P = 42 cm
In △ABC, m∠A=45°. The altitude divides side AB into two parts of 20 and 21 units. Find BC.
Answer:
29 units
Step-by-step explanation:
BC is a side of ACB, which is a 45 45 90 triangle. BC = AB/sqrt2
If In △ABC, m∠A=45°. The altitude divides side AB into two parts of 20 and 21 units. Then BC is 29 units
What is Trigonometry?Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles.
Given,
In △ABC, m∠A=45°. The altitude divides side AB into two parts of 20 and 21 units.
We need to find BC
In triangle ACD,
tan 45° = CD/AD
CD = tan 45° x AD
= 1 x 20= 20 units
In triangle CDB,
tanФ = CD/BD
Ф = tan⁻¹(CD/BD)
= tan⁻¹(20/21)
= 43.6°
so, sin 43.6° = CD/BC
BC = CD/sin 43.6°
= 20/0.689
= 29 units
Hence the length of BC will be 29 units.
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lana 15:02If two events A and B are independent and you know that P(A) = 0.3, what is the value of P(A|B)?
Since the events are independent, we have the following property:
[tex]P(A)=P(A|B)[/tex]That is, the probability of A is the same as the probability of A given B (since the events are independent, event B does not affect event A).
So, if P(A) = 0.3, therefore P(A|B) is also equal to 0.3.
(06.03 MC) Use the expression 5(6 + 4x) to answer the following: Part A: Describe the two factors in this expression. (4 points) Part B: How many terms are in each factor of this expression? Part C: What is the coefficient of the variable term? (2 points)
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data
Expression:
5(6 + 4x)
Step 02:
5(6 + 4x)
A.
5 = factor 1
(6 + 4x) = factor 2
B.
5 = factor 1 ( 1 term)
(6 + 4x) = factor 2 (2 terms)
C.
variable term: 4x
coefficient = 4
That is the full solution.
Which graph represents the table below?
Answer:
The answer will be D because you have to look very close and make sure the 1 is on the x-intercept and not the y-intercept.
The following triangles are not similar. Determine the ratio between AMCD and AOLN. Howcould you change the measurement(s) to make them similar?
The ratio between MD and DC IS
[tex]\frac{MD}{DC}=\frac{14}{6}=\frac{7}{3}[/tex]The ratio of ON to LN
[tex]\frac{ON}{LN}=\frac{49}{21}=\frac{7}{3}[/tex]The ratio CM to CD
[tex]\frac{CM}{CD}=\frac{12}{6}=\frac{2}{1}[/tex]The ratio of OL to LN
[tex]\frac{OL}{LN}=\frac{40}{21}[/tex]To make the
Factor 12x² + 19x - 21.O (6x + 7)(2x − 3)—O (4x - 3)(3x + 7)O(6x-7)(2x + 3)(4x + 3)(3x7)
given the expression
[tex]12x^2+19x-21[/tex]we are loking 2 numbers whose multiplication is equal to 12
and other 2 number whose multiplication is equal to 21
the sum of the cross multiplication is equal of 19, as follows
[tex](4x*3x)+((4x*7)+(3x*-3))+(-3*7)[/tex]factor is
[tex]\left(4x-3\right)\left(3x+7\right)[/tex]correct answer option B
How would you use the Pythagorean Theorem to find the missing length in the triangle shown? Find the missing length.
The given triangle is a right angle triangle. The pythagorean theorem is expressed as
hypotenuse^2 = one leg^2 + other leg^2
From the diagram,
hypotenuse = AB = c
one leg = BC = 9
other leg = AC = 12
By applying the pythagorean theorem, we have
[tex]\begin{gathered} c^2=9^2+12^2\text{ = 81 + 144} \\ c^2\text{ = 225} \\ c\text{ = }\sqrt[]{225} \\ c\text{ = 15} \end{gathered}[/tex]The missing length is 15 cm
You have 4/5 of a pizza left over from your pool party. If you sent 4/9 of the leftover pizza home with your friends how much of the pizza do you have left in the box
If I sent 4/9 of the leftover pizza home with your friends the pizza do I have left in the box is 16/45.
What is pizza?
Pizza is an Italian food consisting of a typically flat, spherical foundation composed of leavened wheat dough that is topped with cheese, tomatoes, and frequently a number of additional toppings. Following that, the pizza is baked at a high temperature, typically in a wood-fired oven. A small pizza is also known as a pizzetta. A pizza maker is referred to as a pizzaiolo.
To get the quantity of the leftover pizza, we need to subtract the pizza sent by me to home from the total pizza I had in my lunch box.
So,
Pizza I have left = Pizza in lunch box - Pizza I have sent
Pizza I have left = 4/5 - 4/9
Pizza I have left = (4(9) - 4(5))/45
Therefore, Pizza I have left is 16/45.
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it snowed 20 inches in 10 days in Montreal. Find the unit rate.
the expression is
[tex]\frac{20}{10}[/tex]We must divide each value by the value of the denominator to obtain the unit ratio
so
[tex]\frac{\frac{20}{10}}{\frac{10}{10}}=\frac{2}{1}=2[/tex]the unit ratio is 2 inches per day
If triangular pyramid P and triangular pyramid D are similar, which of the following statements must be true?
When two figures are similar, the relation or proportion between same measures is the same. So answer is option d.
Abby scored 88, 91, 95, and 89 on her first four history quizzes. What score does Abby need to get on her fifth quiz to have an average of exactly 90 on her history quizzes? a.85b.86c.87a.88
Solution
For this case we can use the definition of average given by:
[tex]\text{Mean}=\frac{x_1+x_2+x_3+x_4+x_5}{5}[/tex]The final score needs to be 90 so we can do this:
[tex]90=\frac{88+91+95+89+x_5}{5}[/tex]And solving for x5 we got:
5*90 = 88+91+95+89+ x5
x5= 450 - 88- 91- 95 -89 = 87
Final answer:
c.87
Basically, the question is asking to solve a problem about rectangular prisms. It says the the shape of one box, with (h) height in feet, has a volume defined by the function:V(h) = h (h-5)(h-6)It says to graph the function. What is the maximum volume for the domain 0
The function representing the volume of the rectangular prism is given to be:
[tex]V(h)=h(h-5)(h-6)[/tex]Since we are expected to find the volume using the graph, we can prepare a table of values for the function using values of h as integers from 1 - 5, such that:
[tex]\begin{gathered} At\text{ }h=1 \\ V(1)=1(1-5)(1-6)=-4\times-5=20 \end{gathered}[/tex]The completed table is shown below:
Hence, we can plot these points on a graph using a graphing calculator for ease of work. This is shown below:
The maximum volume of the prism is represented by the highest point on the graph. The graph's highest point is at:
[tex]h=1.811[/tex]The corresponding value for the volume as can be seen on the graph is:
[tex]V=24.193[/tex]This is the maximum volume of the prism.
To the nearest cubic foot, the maximum volume of the rectangular prism is 24 cubic feet.
Calculate the area of the circle. Round decimal answer to the nearest tenth.
Give the radius of a circle, r, we can find its area by using:
[tex]A=\pi r^2[/tex]In the picture, the 30 ft segment passes from on side of the circle to the other passing thourhg tht center, so it is the diameter. The radius is half the diameter, so:
[tex]r=\frac{30}{2}=15[/tex]Now, we can use the formula for the area to find it:
[tex]A=\pi(15)^2=3.14159\ldots\cdot225=706.8583\ldots\cong706.9[/tex]So, the area is approximately 706.9 ft².
Does the relation in the table represent direct variation, inverse variation, or neither? If it is direct or inverse variation, write an equation to represent the relation. Explain your answer.See image
Statement Problem: Does the relation in the table represent direct variation, inverse variation, or neither? If it is direct or inverse variation, write an equation to represent the relation. Explain your answer.
Solution;
We observe that as the value of x is increasing, the value of y is decreasing. Hence, it has a feature of an inverse variation.
When two variables are inversely related;
[tex]\begin{gathered} x\propto\frac{1}{y} \\ x=\frac{k}{y} \end{gathered}[/tex]But;
[tex]x=5,y=2[/tex]Thus, we have;
[tex]\begin{gathered} x=\frac{k}{y} \\ 5=\frac{k}{2} \\ k=5\times2 \\ k=10 \end{gathered}[/tex]Thus, the equation to represent the information is;
[tex]\begin{gathered} x=\frac{k}{y} \\ \text{Put the value of k in the equation;} \\ x=\frac{10}{y} \end{gathered}[/tex]The equation to represent the information is;
[tex]x=\frac{10}{y}[/tex]The Wong family and the Nguyen family each used their sprinklers last summer. The water output rate for the Wong familys sprinkler was 35 L per hour. The water output for the Nguyen familys sprinkler was 20 L per hour. The families used their spirit for a combined total of 75 hours, resulting in a total water output of 2100 L. How long was each sprinkler used?
Let W be the time of the wong family and N the time of the Nguyen family. We are told that the output rate of the Wong family is 35 L/h and the output for the Nguyen family is 20 L/h, if the total water output is 2100 L, then we can write this mathematically as:
[tex]35W+20N=2100,(1)[/tex]Where "35W" is the total water output of the wong family and "20N" is the total water output of the Nguyen family. The two outputs combined must be 2100. We are also told that the total time is 75 hours, therefore we have:
[tex]W+N=75,(2)[/tex]We get a system of two equations and two variables. We can solve for "W" in equation (2), by subtracting "N" from both sides:
[tex]W=75-N[/tex]Now we can replace this value in equation (1):
[tex]35(75-N)+20N=2100[/tex]Now we apply the distributive property:
[tex]2625-35N+20N=2100[/tex]Now we add like terms;
[tex]2625-15N=2100[/tex]Now we subtract 2625 from both sides:
[tex]\begin{gathered} -15N=2100-2625 \\ -15N=-525 \end{gathered}[/tex]Now we divide both sides by 15:
[tex]N=-\frac{525}{-15}=35[/tex]Now we replace this value in equation (2) where we already solved for W:
[tex]\begin{gathered} W=75-35 \\ W=40 \end{gathered}[/tex]Therefore, the time for the Wong family is 40 hours and the Nguyen family is 35 hours.
ZA and ZB are supplementary angles. If mZA= (8x – 27) and m ZB = (4x + 3), then find the measure of ZA.
Supplementary angles sum up to 180 degrees.
Since mZA and mZB are given to be (8x - 27) and (4x + 3) respectively, we need to know the value of x to be able to find
Supplementary angles sum up to 180 degrees.
Since mZA and mZB are given to be (8x - 27) and (4x + 3) respectively, we need to know the value of x to be able to find
Picture explains it all
Answer: .1$ so 10cents
Step-by-step explanation:
on a horizontal line segment, point A is located at 21, point b is located at 66. point p is a point that divides segment ab in a ratio of 3:2 from a to b where is point p located
We have a one-dimensional horizontal line segment. Three points are indicated on the line as follows:
In the above sketch we have first denoted a reference point at the extreme left hand as ( Ref = 0 ). This is classified as the origin. The point ( A ) is located on the same line and is at a distance of ( 21 units ) from Reference ( Ref ). The point ( B ) is located on the same line and is at a distance of ( 66 units ) from Reference ( Ref ).
The point is located on the line segment ( AB ) in such a way that it given as ratio of length of line segment ( AB ). The ratio of point ( P ) from point ( A ) and from ( P ) to ( B ) is given as:
[tex]\textcolor{#FF7968}{\frac{AP}{PB}}\text{\textcolor{#FF7968}{ = }}\textcolor{#FF7968}{\frac{3}{2}\ldots}\text{\textcolor{#FF7968}{ Eq1}}[/tex]The length of line segment ( AB ) can be calculated as follows:
[tex]\begin{gathered} AB\text{ = OB - OA } \\ AB\text{ = ( 66 ) - ( 21 ) } \\ \textcolor{#FF7968}{AB}\text{\textcolor{#FF7968}{ = 45 units}} \end{gathered}[/tex]We can form a relation for the line segment ( AB ) in terms of segments related to point ( P ) as follows:
[tex]\begin{gathered} \textcolor{#FF7968}{AB}\text{\textcolor{#FF7968}{ = AP + PB }}\textcolor{#FF7968}{\ldots}\text{\textcolor{#FF7968}{ Eq2}} \\ \end{gathered}[/tex]We were given a ratio of line segments as ( Eq1 ) and we developed an equation relating the entire line segment ( AB ) in terms two smaller line segments as ( Eq2 ).
We have two equation that we can solve simultaneously:
[tex]\begin{gathered} \textcolor{#FF7968}{\frac{AP}{PB}}\text{\textcolor{#FF7968}{ = }}\textcolor{#FF7968}{\frac{3}{2\text{ }}\ldots}\text{\textcolor{#FF7968}{ Eq1}} \\ \textcolor{#FF7968}{AB}\text{\textcolor{#FF7968}{ = AP + PB }}\textcolor{#FF7968}{\ldots Eq2} \end{gathered}[/tex]Step 1: Use Eq1 and express AP in terms of PB.
[tex]AP\text{ = }\frac{3}{2}\cdot PB[/tex]Step 2: Substitute ( AP ) in terms of ( PB ) into Eq2
[tex]AB\text{ = }\frac{3}{2}\cdot PB\text{ + PB}[/tex]We already determined the length of the line segment ( AB ). Substitute the value in the above expression and solve for ( PB ).
Step 3: Solve for PB
[tex]\begin{gathered} 45\text{ = }\frac{5}{2}\cdot PB \\ \textcolor{#FF7968}{PB}\text{\textcolor{#FF7968}{ = 18 units}} \end{gathered}[/tex]Step 4: Solve for AP
[tex]\begin{gathered} AP\text{ = }\frac{3}{2}\cdot\text{ ( 18 )} \\ \textcolor{#FF7968}{AP}\text{\textcolor{#FF7968}{ = 27 units}} \end{gathered}[/tex]Step 5: Locate the point ( P )
All the points on the line segment are located with respect to the Reference of origin ( Ref = 0 ). We will also express the position of point ( P ).
Taking a look at point ( P ) in the diagram given initially we can augment two line segments ( OA and AP ) as follows:
[tex]\begin{gathered} OP\text{ = OA + AP} \\ OP\text{ = 21 + 27} \\ \textcolor{#FF7968}{OP}\text{\textcolor{#FF7968}{ = 48 units}} \end{gathered}[/tex]The point ( P ) is located at.
Answer:
[tex]\textcolor{#FF7968}{48}\text{\textcolor{#FF7968}{ }}[/tex]
Finding a final amount in a word problem on exponential growth or decay
Answer:
19g
Explanation:
Given the starting amount as 155grams, the below amount will be left after the 1st half life;
[tex]\frac{155g}{2}=77.5g[/tex]After the 2nd half-life, the below amount will remain;
[tex]\frac{77.5g}{2}=38.75g[/tex]After the 3rd half-life, the below amount will remain;
[tex]\begin{gathered} \frac{38.75}{2}=19.375g\approx19g\text{ (rounding to the nearest gram)} \\ \end{gathered}[/tex]Which graph represents an exponential function
This is algebra two graphing exponential functions
Answer: The Curved Line on Top
Step-by-step explanation: A positive-valued function of a real variable. So the top one
Just learned about this in Algebra 1 about 4 days ago.
This relation map is the musician to the instrument they play. is this relation a function?
Could you help me with this please is from apex
Answer:
Completing the table we have;
Explanation:
Given the table in the attached image, we want to complete the table;
[tex]\text{Interest is 1\% compounded monthly}[/tex]For period 1;
simple interest;
[tex]i_1=Prt=100\times0.01\times1=\text{ \$1.00}[/tex]Compound interest;
[tex]\begin{gathered} f_1=P(1+\frac{r}{n})^{nt}=100(1+\frac{1}{12})^{1(12)}=\text{ \$}101.00 \\ \text{ Interest = }101.00-100=\text{ \$1.00} \end{gathered}[/tex]For period 2;
simple interest;
[tex]i_2=Prt=100\times0.01\times1=\text{ \$1.00}[/tex]compound interest;
[tex]\begin{gathered} f_2=P(1+\frac{r}{n})^{nt} \\ P=f_1=101.00 \\ =101(1+\frac{1}{12})^{1(12)}=\text{ \$}102.01 \\ \text{Interest}=102.01-101=\text{ \$}1.01 \end{gathered}[/tex]Total interest
simple interest;
[tex]i_t=i_1+i_2=1+1=\text{ \$2.00}[/tex]Compound Interest;
[tex]\text{ Total interest}=1.00+1.01=\text{ \$2.01}[/tex]Therefore, completing the table we have;